(a) Copy and complete the table of values, correct to one decimal place, for the relation \(y = 3\sin x + 2\cos x\) for \(0° \leq x \leq 360°\).
x
0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
360°
y
3.0
1.6
-2.0
-3.6
-3.0
2.0
(b) Using scales of 2cm to 30°mon the x- axis and 2cm to 1 unit on the y- axis, draw the graph of the relation \(y = 3\sin x + 2\cos x\) for \(0°\leq x \leq 360°\). (c) Use the graph to solve : (i) \(3\sin x + 2\cos x = 0\) (ii) \(2 + 2\cos x + 3\sin x = 0\).
Explanation
(a)
x
0°
30°
60°
90°
120°
150°
180°
210°
240°
270°
300°
330°
360°
y
2.0
3.2
3.6
3.0
1.6
-0.2
-2.0
-3.2
-3.6
-3.0
-1.6
0.2
2.0
(b) (c)(i) The equation, \(3\sin x + 2\cos x = 0\) has solution where the curve cuts the x- axis, i.e. at A(x = 147°) and B(x = 325.5°). (ii)First, rearrange \(2 + 2\cos x + 3\sin x = 0\) to have the main graph content \(3\sin x + 2\cos x\) on one side of the equation; i.e. \(3\sin x + 2\cos x = -2\). Add the line \(y = -2\) to the graph. The line intersects the curve at the points (x = 180°) and (x = 292.5°). Hence, these are the solutions.
